Homework questions, tension, acceleration, spring scale?

Question

1.Two objects with masses of 3.10 kg and 4.80 kg are connected by a light string that passes over a light frictionless pulley to form an Atwood machine

(a) Determine the tension in the string.
(b) Determine the acceleration of each object.
(c) Determine the distance each object will move in the first second of motion if they start from rest.

2. A 76.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.57 m/s in 1.40 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.80 s and comes to rest.

(a) What does the spring scale register before the elevator starts to move?
(b) What does it register during the first 1.40 s?
(c) What does it register while the elevator is traveling at constant speed?
(d) What does it register during the time it is slowing down?



Thank you sooo much, these are my last two and i just cant get them

Answer 1

The tension in the string is the sum of the forces from the masses, or m1*g+m2*g.

The forces on m2 are its weight m2*g pulling down, and m1*g pulling up. The difference accelerates the mass. The difference is the net force, (m2-m1)*g, the acceleration resulting from that force on m2 is a=f/m2 = (m1+m2)*g/m2. Since the masses move together, the acceleration values are the same, but opposite in sign (one goes up, the other goes down).

Under constant acceleration the distance traveled is s = .5*a*t^2. Plug in a from above and the time of one second for t.

The spring scale registers the man's weight of 76k*gN When the elevator is at rest. (g is the acceleration due to gravity = 9.8m/sec^2)

The acceleration of the elevator adds to the force of gravity when it is rising. The elevator acceleration is vmax/tup (assuming constant acceleration). (vmax = 1.57m/s, tup = 1.40s) So the force on the scale will be m*(g+vmax/tup).

At constant speed there is no acceleration of the elevator, and the scale reads the same as when it was at rest.

The downward acceleration is vmax/tdown. (tdown = 1.80s). This subracts from the force of gravity, so that the weight of the man will be m*(g - vmax/tdown)

Answer 2

to confirm the photograph could be good. i will make a danger re the association based on your "40N. unbelievable from each and every cable". If the two the lots are unbelievable, e.g. the cables run around pulleys and the size is horizontal between the pulleys,and you're on earth, the rigidity interior the cables would be 40N. the explanation is that in case you bumped off one 40N. mass and linked its cable to a fastened element as a replace, the size could examine 40N. One mass balances the different.